Wikipedia:Reference desk/Archives/Mathematics/2009 April 11
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April 11
editLimit involving moment generating functions
editHi there - I'm looking to prove the result that if X is a real R.V. with moment generating function , then if the MGF is finite for some theta > 0, . I really have absolutely no clue where to start - is it central limit theorem related? I could really use a hand on this one, thanks a lot!
Otherlobby17 (talk) 06:51, 11 April 2009 (UTC)
- Look at the integral that defines , namely . If it is finite, then , which is stronger than you need. McKay (talk) 09:14, 11 April 2009 (UTC)
- Fantastic, thanks very much buddy :) Otherlobby17 (talk) 04:15, 12 April 2009 (UTC)
Graph
editIs it possible for the graph of a continuous function to touch the x-axis without there being a repeated root? 92.0.38.75 (talk) 13:02, 11 April 2009 (UTC)
- Are you referring to polynomial functions, rather than the less restrictive class of continuous functions? It may be my own ignorance, but I don't know the definition of a repeated root for an arbitrary continuous function. But in the case of a polynomial function, any instance where the function touches but does not pass through the x-axis must be a repeated root. To see this (assuming you have had calculus), note that the derivative at the root must be zero for the curve to just barely kiss the graph. Take the derivative first:
- If the root is , the term appears in every product except one (when b = i). Now set , as well as setting the expression equal to zero (knowing that the derivative is zero at the root); all that is left is
- Ergo there must be another term , or equivalently, the multiplicity of the root is greater than one.
- Hope this helps, --TeaDrinker (talk) 15:01, 11 April 2009 (UTC)
What type of root does the absolute value function |x| have at 0? What about at 0? I think the concept "repeated root" is only usually defined for functions with derivatives. The order of the root is the number of values f(x), f'(x), f"(x),... which are zero. McKay (talk) 01:50, 12 April 2009 (UTC)
- You could extend that to say that if a function is continuous but not differentiable at a zero, then the zero has order 0, analogously to the way one sometimes thinks of continuous functions as being 0-times differentiable, or C0. I don't know if this is useful for anything though. Algebraist 11:44, 12 April 2009 (UTC)
What symbol is this?
editRan across this symbol in a mathematics book but I have no idea what it is, or how to type it up in LaTeX. It's a bit like a cursive capital X with a short horizontal line through the middle. My first instinct was but it looks nothing like it. 128.86.152.139 (talk) 14:17, 11 April 2009 (UTC)
- Hmm, I don't know anything fitting that description, but have you tried the comprehensive LaTeX symbols list? --TeaDrinker (talk) 14:23, 11 April 2009 (UTC)
- I'm going through it now. :( But it's incredibly long... 128.86.152.139 (talk) 14:36, 11 April 2009 (UTC)
- What book is it? What's the context in which the symbol was used? --TeaDrinker (talk) 14:45, 11 April 2009 (UTC)
- I'm going through it now. :( But it's incredibly long... 128.86.152.139 (talk) 14:36, 11 April 2009 (UTC)
- If you'd have said a vertical line, I'd have said Zhe. It's a shame you didn't. 163.1.176.253 (talk) 14:57, 11 April 2009 (UTC)
don't forget that a lot of people cross weird things. a close acquaintance crosses their v's! Maybe it's just a fancy x, so that you don't think it's a normal x? :) 79.122.103.33 (talk) 15:49, 11 April 2009 (UTC)
- Do you mean ? That's just a fraktur capital X. Algebraist 16:37, 11 April 2009 (UTC)
- Brilliant, thanks. For context, my lecturer uses it to denote a set of data: Is there a non-Blackletter-style font, though? 128.86.152.139 (talk) 02:34, 12 April 2009 (UTC)
how does variable-length testing work?
editlet's say I buy a rigged coin (slightly favors one side) but forget what side is favored
could i just write a script that i put throws into one after the other, and at each stage it tries that many with a fair coin (for example at throw #5 it throws a fair coin five times) but REPEATEDLY, a MILLION times, to see how many times the fair coin behaves the same way under that many throws, ie as a percentage.
Then if the fair coin only behaves that way 4% of the million times, then it would be 96% confident that the currently winning side is weighted?
here are real examples i just ran with a script: if at throw #10 the count is 7 heads to 3 tails (70% heads), it ran a million times ten throws and came up in 172331 of them (17%) with at least that many. So it would report 83% confidence that heads are weighted.
if at throw #50 the count is 35 heads to 15 tails (70% heads), it ran a million times fifty throws and came up in 3356 of them (0.33%) with at least that many. So it would report report 99.67% confidence heads are weighted.
#1: t
0 head 1 tails
50% conf. heads weighted
#2: t
0 heads 2 tails
50% conf. heads weighted
#3: h
1 head 2 tails
50% conf. heads weighted
...
#10: h
7 heads 3 tails
83% conf. heads weighted
...
#50:h
35 heads 15 tails
99.7% conf. heads weighted
is that really how statistics works? if I write my script like I intend to will it be accurate? Also how many decimal places should I show if I am running the 'monte carlo' simulation with a million throws?
Is a million throws accurate enough to see how often a fair coin behaves that way, or should I up it to a billion or even more? could i use a formula, and if so what? (i dont know stats).
Thanks! 79.122.103.33 (talk) 15:30, 11 April 2009 (UTC)
- The confidence interval is much tighter than that, see binomial distribution. 66.127.52.118 (talk) 20:13, 11 April 2009 (UTC)
why does modulus of a random integer in a high range favor lower answers SO SLIGHTLY?
editsay rand() returns 0-32767 but you want 0-100 - you can just do rand() % 100 which is pretty much an accepted programming practice but results in a very slighly skewed distribution.
My question is, how come the skew in the distribution is so incredibly slight?? Here I did it a million times:
0:10137 1:9967 2:10225 3:10157 4:9921 5:10096 6:10087 7:9924 8:9876 9:9994 10:10052 11:10022 12:10098 13:9940 14:10080 15:9939 16:9967 17:10067 18:9930 19:10058 20:10072 21:9882 22:9940 23:9793 24:10051 25:10105 26:10079 27:9970 28:9998 29:10197 30:9868 31:9979 32:10006 33:10014 34:9991 35:10062 36:9641 37:10054 38:9938 39:10221 40:9957 41:10064 42:9913 43:9858 44:10050 45:10080 46:10010 47:10009 48:10147 49:9971 50:10107 51:10083 52:9943 53:9998 54:9926 55:10036 56:9965 57:10048 58:10130 59:10049 60:9889 61:9843 62:10067 63:9918 64:10109 65:10201 66:10037 67:10049 68:9940 69:10011 70:10061 71:9946 72:10017 73:9781 74:9946 75:9986 76:10180 77:9888 78:9850 79:10034 80:10186 81:9803 82:9948 83:10040 84:9984 85:10109 86:9986 87:10006 88:9883 89:9834 90:9921 91:10002 92:10191 93:10091 94:9990 95:9910 96:9837 97:9793 98:10097 99:9894
I barely see the effect unless I know to look for it (at 0 does come up more than 99, but then again 1 doesn't...)
Here they are as percentages of 10,000 (the actual expected number):
101.37% 99.67% 102.25% 101.57% 99.21% 100.96% 100.87% 99.24% 98.76% 99.94% 100.52% 100.22% 100.98% 99.4% 100.8% 99.39% 99.67% 100.67% 99.3% 100.58% 100.72% 98.82% 99.4% 97.93% 100.51% 101.05% 100.79% 99.7% 99.98% 101.97% 98.68% 99.79% 100.06% 100.14% 99.91% 100.62% 96.41% 100.54% 99.38% 102.21% 99.57% 100.64% 99.13% 98.58% 100.5% 100.8% 100.1% 100.09% 101.47% 99.71% 101.07% 100.83% 99.43% 99.98% 99.26% 100.36% 99.65% 100.48% 101.3% 100.49% 98.89% 98.43% 100.67% 99.18% 101.09% 102.01% 100.37% 100.49% 99.4% 100.11% 100.61% 99.46% 100.17% 97.81% 99.46% 99.86% 101.8% 98.88% 98.5% 100.34% 101.86% 98.03% 99.48% 100.4% 99.84% 101.09% 99.86% 100.06% 98.83% 98.34% 99.21% 100.02% 101.91% 100.91% 99.9% 99.1% 98.37% 97.93% 100.97% 98.94%
As you can see they're all over the place.
So I'll do it a billion times:
0:10038140 1:10008197 2:10009360 3:10006955 4:10011825 5:10010609 6:10009413 7:10006938 8:10011526 9:10010894 10:10010597 11:10009374 12:10009683 13:10007576 14:10011881 15:10009578 16:10010504 17:10009339 18:10009367 19:10010843 20:10006451 21:10006077 22:10009165 23:10014474 24:10006321 25:10006088 26:10007508 27:10007083 28:10008172 29:10009126 30:10011141 31:10011209 32:10009601 33:10011616 34:10006668 35:10008558 36:10012031 37:10011200 38:10008657 39:10011348 40:10012982 41:10012670 42:10011145 43:10008010 44:10011152 45:10009978 46:10011937 47:10010535 48:10008799 49:10006801 50:10009905 51:10009997 52:10007276 53:10012822 54:10012214 55:10005860 56:10010537 57:10010839 58:10008926 59:10011667 60:10008250 61:10012131 62:10003874 63:10005923 64:10014245 65:10009392 66:10009417 67:9982730 68:9978860 69:9980179 70:9978155 71:9982744 72:9977599 73:9976077 74:9981662 75:9977978 76:9982794 77:9981410 78:9982701 79:9978788 80:9977564 81:9980187 82:9980063 83:9976760 84:9980559 85:9978017 86:9980910 87:9981715 88:9978261 89:9981133 90:9979202 91:9976322 92:9977249 93:9976058 94:9977878 95:9984202 96:9980344 97:9981362 98:9978432 99:9979728
the effect becomes clear... however it is TINY!!!
Here they are as percentages (of the expected 10million):
100.3814% 100.08197% 100.0936% 100.06955% 100.11825% 100.10609% 100.09413% 100.06938% 100.11526% 100.10894% 100.10597% 100.09374% 100.09683% 100.07576% 100.11881% 100.09578% 100.10504% 100.09339% 100.09367% 100.10843% 100.06451% 100.06077% 100.09165% 100.14474% 100.06321% 100.06088% 100.07508% 100.07083% 100.08172% 100.09126% 100.11141% 100.11209% 100.09601% 100.11616% 100.06668% 100.08558% 100.12031% 100.112% 100.08657% 100.11348% 100.12982% 100.1267% 100.11145% 100.0801% 100.11152% 100.09978% 100.11937% 100.10535% 100.08799% 100.06801% 100.09905% 100.09997% 100.07276% 100.12822% 100.12214% 100.0586% 100.10537% 100.10839% 100.08926% 100.11667% 100.0825% 100.12131% 100.03874% 100.05923% 100.14245% 100.09392% 100.09417% 99.8273% 99.7886% 99.80179% 99.78155% 99.82744% 99.77599% 99.76077% 99.81662% 99.77978% 99.82794% 99.8141% 99.82701% 99.78788% 99.77564% 99.80187% 99.80063% 99.7676% 99.80559% 99.78017% 99.8091% 99.81715% 99.78261% 99.81133% 99.79202% 99.76322% 99.77249% 99.76058% 99.77878% 99.84202% 99.80344% 99.81362% 99.78432% 99.79728%ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Now the effect becomes nice and clear. (But is still tiny)
My questions are:
- why did I have to do a BILLION (which is a huge number) times to see this nice and clear pattern?
- why is the pattern so TINY
- why does the switch from over the expected (were it an even distribution) to under the expected CLEARLY happen at 66? wouldn't 50 make more sense? It's a huge and clear shift and I bet it would repeat around there, given that the numbers above and below are clearly all in line and it's not like a statistical fluke (e.g. if 64 through 67, were just a hair away from each other and not actually in order but 66 "happened to" win out) , it follows the pattern decisively... I wonder why 66 (or two thirds of 99) is where it happens, it seems odd... why?
So: what are the mathematical reasons for such a tiny tiny slight favoring (like 0.1% it seems like over a billion iterations of the lower modulus numbers?), why don't they show up over a million iterations, and why does the shift from over the expected to under the expected number happen at 66 of 99 instead of something sensible like 50?
Thank you! 79.122.103.33 (talk) 17:43, 11 April 2009 (UTC)
- I'm not quite sure I understand the problem. The remainders from 0 through 67 occur 328 times in the range, while the remainders from 68 through 99 only occur 327 times. So we would expect any low number (0–67) to occur with probability 328/32768 = 1.001% and any high number (68–99) with probability 327/32768 = 0.998%. It easily drowns in sheer randomness for too few trials. I have no idea why your result seems to treat 67 as a high number though. That seems weird. —JAO • T • C 18:03, 11 April 2009 (UTC)
- Jao's answer is correct. For another explanation of why the shift happens where it does, think of a system where rand() returns numbers from 0-5 (like a die). Now, if you wanted a number between 0 and 4 exclusive, you could do rand() % 4, with the following possible outcomes:
- rand() returns 0 - result 0
- rand() returns 1 - result 1
- rand() returns 2 - result 2
- rand() returns 3 - result 3
- rand() returns 4 - result 0
- rand() returns 5 - result 1
- It is easy to see that 0 or 1 would occur twice as often as the other numbers. If you made a similar table for 0-32767 and 0-99, you would see that the numbers 0-67 occur once more than the other numbers (namely, when the generator returns a number between 32000 and 32767). decltype (talk) 18:43, 11 April 2009 (UTC)
- thanks for the answers! They make sense. So how would you correctly choose a number between 0 and 3 inclusive using a generator that goes from 0-5 inclusive, in an equally distributed way? Should you just reflip on a 4 or 5, no matter how many times you keep getting it? 79.122.103.33 (talk) 21:25, 11 April 2009 (UTC)
- If you want an algorithm that's sure to stop, you can just flip it twice and see which ordered pair (like (1,1), (0,4) or (5,1)) you get. As there are 36 such ordered pairs, any scheme that maps nine of them to 0, nine to 1, nine to 2 and nine to 3 works. —JAO • T • C 22:12, 11 April 2009 (UTC)
To get uniformity at little cost, just reject values of rand() greater than 32699. Incidentally, it is usually recommended to use division rather than mod for this problem. That is, instead of rand() % 100, use rand() / 327 (after rejecting values above 32699). If rand() was perfectly random, it wouldn't make any difference, but division is considered less likely to magnify the imperfections of imperfect random number generators. McKay (talk) 01:45, 12 April 2009 (UTC)
okay then what is the answer
editthis is in relation to my thread two above (about weighted coins). so how many flips out of how many would I need to reach to be sure my coin isn't fair, for 75%, 90%, 95%, 98.5%, 99%, 99.9% confidence...
what is the formula? (this is not homework) 79.122.103.33 (talk) 21:21, 11 April 2009 (UTC)
- The number of flips of the weighted coin that are necessary to ascertain which direction the weight is in (or to ascertain that it is weighted at all) to a specified level of confidence depends on the extent of the weight. At a fixed level of confidence, a coin with 2/3 probability of landing heads will be determined to be weighted much sooner than a coin with 501/1000 probability of landing heads. So the answer depends upon the amount of skew. Eric. 131.215.159.99 (talk) 23:32, 11 April 2009 (UTC)
how to do monte carlo properly
editif after doing n flips and getting a certain number of heads, I want to be exactly 95% sure that the results show my coin favors heads (isnt fair) but I'm really bad at statistics, and want to do monte carlo instead, could I see what the most heads is in 20 runs (20 because 19/20th is 95%) by making a list of "most heads out of n flips in 20 runs" a billion times, average those numbers, and get my 95% threshold for n?
For example, if I want to see what number out of 50 flips my coin has to beat for me to be exactly 95% sure that it isn't fair, I average a billion metaruns of "most heads out of 50 flips, in 20 tries"?
sorry that I'm such a zero in statistics. this must be so frustrating to anyone who actually knows what's going on. anyway does this proposed methodology work for my intended confidence level (95%)?
if not this then what is the correct monte carlo method for the 95% interval? Thanks!79.122.103.33 (talk) 21:45, 11 April 2009 (UTC)
- The distribution of the number of head in 20 tosses of a fair coin is approximately a normal distribution with a mean of 10 and a variance of 5, so a standard deviation of sqrt(5). In a normal distribution, 95% of observations are less than 1.65 standard deviations above the mean (since your question is "does my coin favour heads" then this is a one-sided test). 10 + 1.65 x sqrt(5) is approximately 13.7. So the probability that a coin that does not favour heads will score more than 13 heads in 20 tosses is less than 5%. So if your coin scores 14 or more heads, you can be "95% certain" that it favours heads. See hypothesis testing and Z-test for more details. Gandalf61 (talk) 12:25, 12 April 2009 (UTC)
- You need to know the prior probability that the coin is biased before you can answer such a question precisely. See hypothesis testing (mentioned by Gandalf) for some discussion. Really, you've asked the question a couple very different ways: 1) you have a coin that is known to be biased, but you're not sure if it's towards heads or towards tails (let's say that's an implicit assumption that it's 50-50 guess between heads-biased or tails-biased); or 2) you have a coin that might be biased (with some unknown probability) and you want to check. Case 1 is straightforward: the null hypothesis is that the coin is biased towards tails, then compare the outcome of your experiment with the binomial or normal distribution. Case 2 is harder: for example, say you have 1000 coins, of which 999 are fair and one is 60-40 biased towards heads. You pick one of those coins uniformly at random, flip 20 times and get 14 heads. Are you 95% confident that you picked the biased coin? Nope, because of the prior probability distribution. In this case you'd use Bayes's theorem in conjunction with the known 0.999 prior probability to interpret the result. But if you don't know the prior probability, it's not so clear what to do. 66.127.52.118 (talk) 12:46, 12 April 2009 (UTC)
- not knowing the prior probabiity, what can one do? If a magician lets you check a coin he's using, what are you supposed to guess for the chances it's fair? How woud you check? (to be 90% / 95% / 98% / 99% etc sure in your conclusion?) Thanks 94.27.222.70 (talk) 22:29, 12 April 2009 (UTC)
- That is to some extent a question about philosophy rather than mathematics. See Bayesian probability and frequency probability (as well as statistical hypothesis testing already mentioned) for some discussion. 66.127.52.118 (talk) 01:19, 13 April 2009 (UTC)